If $t = \sec x$, solve $t^2 + t - 1 = a$ and find the range of values of $a$

Question

If $t = \sec x$, solve $t^2 + t - 1 = a$ and find the range of values of $a$

Answer 1

Hint:

Complete the square: $$t^2+t-1=\Bigl(t+\frac12\Bigr)^{\!2}-\frac14-1,$$ so the equation come down to $$\Bigl(t+\frac12\Bigr)^{\!2}=a+\frac54.$$

Conditions resulting from $t=\sec x$:

As the range of $\sec x$ is $(-\infty,-1]\cup[1,+\infty)$, the range of $t+\frac12$ is $\bigl(-\infty,-\frac12\bigr]\cup\bigl[\frac32,+\infty\bigr)$, so that squaringwe obtain the condition $$a+\frac 54\ge\frac14.$$

Answer 2

$$t^2-t-(a+1)=0$$ Then Quadratic Formula $$t=\frac{1\pm\sqrt{1+4a+4}}{2}=\frac 12\pm\frac 12\sqrt{4a+5}$$

Answer 3

since,

$\sec(x)\in (-\infty,-1]\cup[1,\infty) $

$t\notin(-1,1)$

$ f(t) = t^2+t-1,\\ f'= 2t-1 = 0 \quad at \quad t=-1/2\\f(-1)=-1,\ f(-1/2)=-5/4,\ f(1)=1$

if you look at the graph of f(t), you will find: $ \quad a\in[-1,\infty)$